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An equation is given, followed by one or more roots of the equation. In each case, determine the remaining roots. $$\begin{array}{l}x^{7}-3 x^{6}-4 x^{5}+30 x^{4}+27 x^{3}-13 x^{2}-64 x+26=0 \\\x=3-2 i, x=-1+i, x=1\end{array}$$

Short Answer

Expert verified
The remaining roots are \( x = 2 \) and \( x = -1 \).

Step by step solution

01

Understand Complex Conjugates

Given that two roots are complex numbers, we need to remember that if the coefficients of the polynomial are real, complex roots must come in conjugate pairs. Thus, the roots include \( x = 3-2i \), \( x = 3+2i \), \( x = -1+i \), and \( x = -1-i \).
02

Remaining Known Real Root

We are given that \( x = 1 \) is another root.
03

Identify Unidentified Roots

Since the polynomial is of degree 7, there must be 7 roots altogether. We have so far identified five roots: \( x = 3 \pm 2i \), \( x = -1 \pm i \), and \( x = 1 \). Thus, two roots remain to be identified.
04

Polynomial Division

To determine the remaining roots, perform polynomial division. First, divide the polynomial by \((x - 1)\). The result is a polynomial of degree 6.
05

Factor for Known Complex Pairs

Continue dividing the resulting polynomial by the factor \((x - (3-2i))(x - (3+2i))\) and then by \((x - (-1+i))(x - (-1-i))\), which correspond to known complex conjugate roots.
06

Solve the Remaining Quadratic

After all the divisions, you should be left with a quadratic equation. Solve this quadratic to identify the remaining roots.
07

Check Remaining Roots

The final roots found from solving the quadratic should be \( x = 2 \) and \( x = -1 \). Check these for any computational errors, confirming your result.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Complex Conjugates
In the world of polynomials, complex conjugates are essential, especially when dealing with real coefficients. When we talk about a complex conjugate, we refer to pairs of complex numbers that have the same real part but opposite imaginary parts. For instance, if one root is given as \( x = 3-2i \), then its complex conjugate is \( x = 3+2i \). This concept is crucial in mathematics because when a polynomial has real coefficients, its complex roots must appear in conjugate pairs. This ensures that the polynomial has real coefficients throughout its expansion and simplifies the solving process. It's a handy tool, especially when identifying potential roots in higher-degree polynomials.
Polynomial Division
Polynomial division is the methodical process of breaking down a polynomial to make it more manageable. It works similarly to long division with numbers. In our problem, after identifying known roots, we reduce the polynomial by dividing it by expressions related to these roots.
For example, with a real root of \( x = 1 \), we first divide the entire polynomial \( x^7 - 3x^6 - 4x^5 + 30x^4 + 27x^3 - 13x^2 - 64x + 26 \) by \( (x - 1) \). This simplifies the polynomial to a degree of 6.
Subsequently, we further streamline by dividing this new polynomial by the polynomial formed by our conjugate pairs, \( (x - (3-2i))(x - (3+2i)) \), and \( (x - (-1+i))(x - (-1-i)) \). This step-by-step breakdown, while sometimes tedious, helps us systematically strip down the polynomial to its simplest form.
Quadratic Equation
Once polynomial division simplifies our polynomial to a manageable degree, we might encounter a quadratic equation. A quadratic equation typically has the form \( ax^2 + bx + c = 0 \). Solving a quadratic can unveil the roots that remained elusive during polynomial division.
Methods often used to solve these equations include factoring, completing the square, and using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Our division process should ideally result in a quadratic equation where these methods apply, finally revealing any remaining roots of our original polynomial equation. Understanding and mastering quadratic equations are foundational skills in algebra, vital for tackling higher-degree polynomials.
Real Coefficients
Real coefficients in a polynomial mean that the numbers multiplying the terms of the polynomial are real numbers (as opposed to imaginary numbers). In the context of complex roots, this has significant implications, because, as we previously discussed, any complex roots must appear with their conjugates.
This property not only maintains the integrity of the polynomial in accordance with real number operations but also provides a strategic advantage in solving such equations. When dealing with real coefficients, remember that half of any complex roots you encounter will have their pairs automatically determined, which simplifies part of the solving process by limiting computational guesswork.

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Most popular questions from this chapter

\(\frac{1}{x^{4}+x^{3}+2 x^{2}+x+1}\) Hint: To factor the denominator, replace \(2 x^{2}\) with \(x^{2}+x^{2}\) and group as follows: \(\left(x^{4}+x^{3}+x^{2}\right)+\left(x^{2}+x+1\right)\)

One of the basic properties of real numbers is that if a product is equal to zero, then at least one of the factors is zero. Exercises show that this property also holds for complex numbers. For Exercises 79 and \(80,\) assume that \(z w=0,\) where \(z=a+b i\) and \(w=c+d i\) If \(a \neq 0, \text { prove that } w=0 . \text { (That is, prove that } c=d=0 .)\)

Determine whether the given value is a zero of the function. \(f(x)=x^{3}-3 x^{2}+3 x-3\) (a) \(x=\sqrt[3]{2}-1\) (b) \(x=\sqrt[3]{2}+1\)

Determine the partial fraction decomposition for each of the given rational expressions. Hint: In Exercises \(17,18,\) and \(26,\) use the rational roots theorem to factor the denominator. $$\frac{x^{3}+x-3}{x^{4}-15 x^{3}+75 x^{2}-125 x}$$

Scipio Ferro of Bologna well-nigh thirty years ago discovered this rule and handed it on to Antonio Maria Fior of Venice, whose contest with Niccolò Tartaglia of Brescia gave Niccolò occasion to discover it. He / Tartaglial gave it to me in response to my entreaties, though withholding the demonstration. Armed with this assistance, I sought out its demonstration in /various / forms. - Girolamo Cardano, Ars Magna (Nuremberg, 1545 ) The quotation is from the translation of Ars Magna by T. Richard Witmer (New York: Dover Publications, 1993 ). In his book Ars Magna (The Great Art) the Renaissance mathematician Girolamo Cardano \((1501-1576)\) gave the following formula for a root of the equation \(x^{3}+a x=b\). $$x=\sqrt[3]{\frac{b}{2}+\sqrt{\frac{b^{2}}{4}+\frac{a^{3}}{27}}}-\sqrt[3]{\frac{-b}{2}+\sqrt{\frac{b^{2}}{4}+\frac{a^{3}}{27}}}$$ (a) Use this formula and your calculator to compute a root of the cubic equation \(x^{3}+3 x=76\) (b) Use a graph to check the answer in part (a). That is, graph the function \(y=x^{3}+3 x-76,\) and note the \(x-\) intercept. Also check the answer simply by substituting it in the equation \(x^{3}+3 x=76\)

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